If a 3x3 matrix $A$ has only one real eigenvalue, then we can deduce that the eigenvalue's algebraic multiplicity is either 1 ($A$'s other eigenvalues are complex) or 3 ($A$'s characteristic polynomial is of the form $(\lambda-a)^3$).

Thus, its corresponding eigenspace is 1-dimensional in the former case and either 1, 2 or 3-dimensional in the latter (as the dimension is at least one and at most its algebraic multiplicity).

p.s. The eigenspace is 3-dimensional if and only if $A=kI$ (in which case $k=\lambda$).